a) Assume $A = (a_{ij})_{1 \le i,j \le n}$ and $A^2 = (m_{ij})_{1 \le i,j \le n}$. The property $A^T = -A$ leads to the relations $a_{ji} = -a_{ij}$, for $i, j = 1, \dots, n$, that is $A$ is antisymmetric. Then
$$m_{ii} = \sum_{j=1}^{n} a_{ij} a_{ji} = - \sum_{j=1}^{n} a_{ij}^2, \quad i = 1, \dots, n.$$
If $A^2 = O_n$, then $m_{ii} = 0$, $i = 1, \dots, n$. Since $A \in \mathcal{M}_n(\mathbb{R})$, we obtain $a_{ij} = 0$, for $i, j = 1, \dots, n$. So $A = O_n$.

b) From the assumption, $A = B^*$, where $B^*$ is the adjoint of $B$. Since $n$ is an odd integer number, we obtain $\det(A) = \det(A^T) = \det(-A) = (-1)^n \det(A) = -\det(A)$. Then we obtain $\det(A) = 0$. Therefore $\det(BB^*) = \det(B) \cdot \det(B^*) = \det(B) \cdot \det(A) = 0$. From the relation $BB^* = \det(B)I_n$, we get $\det(B) = 0$. Hence $\rank(B) \le n-1$ and $BB^* = O_n$.

Case 1. If $\rank(B) \le n-2$, then $B^* = O_n$. Thus, $A^2 = (B^*)^2 = O_n$.

Case 2. If $\rank(B) = n-1$, then $B^* \ne O_n$. From the Sylvester rank inequality, we have $\rank(B^*) \le \rank(BB^*) + n - \rank(B) = 1$. We conclude $\rank(B^*) = 1$ and $(B^*)^2 = \tr(B^*)B^*$. But $\tr(B^*) = \tr(A) = 0$ because $a_{ii} = -a_{ii}$, $i = 1, 2, \dots, n$. Finally, we get $A^2 = (B^*)^2 = O_n$.